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In principle, the question seems too elementary for this site, or at least borderline. But, since I gave an answer, I vote to reopen, not to loose the information. In general, note that the eventuality of a unsuited question with a good answer (I'm not claiming this is the case, of course) has been contemplated in this site; consider e.g. the badge "Reversal".
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Pietro MajerNov 21 '11 at 10:38

@Pietro: you could always post your answer to the stackexchange site too.
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George LowtherNov 21 '11 at 13:22

1 Answer
1

The function $f(x)$ is $\mu([0,x])$ where $\mu$ is the radon measure $\sum_{n\in\mathbb{Z} _ +} 2^{-n}\delta _ {q_n}$, and $\mu$ is singular w.r.to the Lebesgue measure $\lambda$ (in fact, $\operatorname{supp}(\mu)=(0,1)\cap\mathbb{Q}$). So the absolutely continuous part $\mu ^ a$ w.r.to $\lambda$ is zero, and the Radon-Nikodym derivative $d\mu ^ a/d \lambda$ is also zero; but this coincides a.e. with the derivative of $f$.
Note that (depending on the particular chosen enumeration of $(0,1)\cap\mathbb{Q}$) there might be infinitely many irrational points $x$ where $f$ is continuous and derivable with any value of $f'(x)$; a Lebesgue null set though.

edit. A more elementary argument. Consider the nested family of open nbd's of $(0,1)\cap\mathbb{Q},$